Question : If $x^4+x^{-4}=47, x>0$, then what is the value of $x+\frac{1}{x}-2?$
Option 1: 1
Option 2: 0
Option 3: 5
Option 4: 3
Correct Answer: 1
Solution :
Given expression,
$x^4+x^{-4}=47$
⇒ $x^4+\frac{1}{x^4}=47$
Adding 2 on both sides,
⇒ $x^4+\frac{1}{x^4}+2=47+2$
⇒ $x^4+\frac{1}{x^4}+2\times x^2\times\frac{1}{x^2}=49$
⇒ $(x^2+\frac{1}{x^2})^2=7^2$ [As $a^2+b^2+2ab=(a+b)^2$]
⇒ $x^2+\frac{1}{x^2}=7$
Again adding 2 on both sides,
⇒ $x^2+\frac{1}{x^2}+2=7+2$
⇒ $x^2+\frac{1}{x^2}+2\times x\times\frac1x=9$
⇒ $(x+\frac1x)^2=3^2$
⇒ $x+\frac1x=3$
⇒ $x+\frac1x-2=3-2$
$\therefore x+\frac1x-2=1$
Hence, the correct answer is 1.
Related Questions
Question : If $x^2+\frac{1}{x^2}=\frac{7}{4}$ for $x>0$; then what is the value of $x+\frac{1}{x}$?
Question : If $x^2+\frac{1}{x^2}=\frac{7}{4}$ for $x>0$, what is the value of $(x^3+\frac{1}{x^3})$?
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